Jotto Logic

Getting a zero word is a good move.

If you can come up with 5 words of 5 letters each, all unique, you greatly narrow the possibilities. Since there are only 5 vowels though this is kinda tough.

Code:

hr    1
flr    1
bland    2
node    1
dane    2
dorky    1
barn    2
bored    3

Since the letters 'stick' don't appear, I've taken the liberty of removing them from the remaining words.

It's not a massive jump from here to assume that R is in the correct solution. flh are not as common letters, and well, if you're wrong, you just re-examine this assumption.
So taking R out, and subtracting 1 from each word containing it, you get:
s, t, i, c, k, f, l, h - eliminated
r - in the word

Code:

band    2
node    1
dane    2
doky    0
ban    1
boed    2

Then doky are obviously not in it either
eliminated: s, t, i, c, k, f, l, h, d, o, k, y
in the solution: r
(unknown: a, b, e, g, j, m, n, p, q, u, v, w, x, z)

Running another pass to remove doky,

Code:

ban    2
ne    1
ane    2
ban    1
be    2

There's a contradiction here: ban 2 - ban 1
Conclusion: R is not in the solution.

Dropping back a bit,
eliminated: s t i c k r

Code:

h    1
fl    1
bland    2
node    1
dane    2
doky    1
ban    2
boed    3

h is obviously in the solution, as is either f or l.
Another less immediate conclusion - ban has 2, bland has 2. So l and d are not in the solution. Going back...
eliminated: s t i c k r l d
in the solution: h, f

Code:

ban    2
noe    1
ane    2
oky    1
ban    2
boe    3

in the solution: h f b o e - again, not consistent.


I think you must still have something wrong with the original numbering, cause as it stands there is no solution, or I slipped up on my logic somewhere.